Q 3.6-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that when Euler’s method is used to approximate the solution of the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{5}\mathbf{y}$ y(0) = 1  , at x = 1, then the approximation with step size h is ${\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{5}\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{h}}}$.

Step-by-Step Solution

Answer

Proved

Step 1: Apply Euler’s formula

Here  $\mathrm{y}\text{'}=5\mathrm{y},\mathrm{y}\left(0\right)=1$

 

Apply the Euler’s formula 

 

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}})$

 

In our case of differential equation our equation is 

 ${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{h}5{\mathrm{y}}_{\mathrm{n}}$

The interval is $\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{1-0}{\mathrm{n}}=\frac{1}{\mathrm{n}}=\mathrm{h}$

Step 2: Substitute in the Euler’s formula

Now

$$\begin{gathered} {\mathrm{y}}_1=1(1+5\mathrm{h}) \\ {\mathrm{y}}_2=1+5\mathrm{h}(1+5\mathrm{h})-(1+5\mathrm{h}{)}^2 \\ . \\ . \\ . \\ {\mathrm{y}}_{\mathrm{n}}=(1+5\mathrm{h}{)}^{\mathrm{n}} \\ {\mathrm{y}}_{\mathrm{n}}=(1+5\mathrm{h}{)}^{\frac{1}{\mathrm{h}}} \end{gathered}$$

Hence, it is proved that ${\mathbf{\text{y}}}_{\mathbf{\text{n}}}{\mathbf{\text{ = (1 + 5h)}}}^{\frac{\mathbf{\text{1}}}{\mathbf{\text{h}}}}$